Chapter 2.6, Problem 19AYU

To determine

To find:

a. To build a model that expresses distance $d$ between two cars as a function of time $t$.

Answer to Problem 19AYU

Solution:

a. $d\left(t\right)=\sqrt{\left(2500t^2-360t+13\right)}$

Explanation of Solution

Given:

Two cars are travelling perpendicular to each other with velocities 30 miles/hr and 40 miles/hr. One is 2 miles south of intersection and other is 3 miles east of intersection.

Calculation:

a. The motion of the cars can be represented as the figure below:

Let, $v_1$ be the velocity of the car travelling south = 30mph.

Let, $v_2$ be the velocity of the car travelling west = 40 mph.

Let, $d_1$ be the distance from the intersection at time $t$ of the car travelling south.

Let, $d_2$ be the distance from the intersection at time $t$ of the car travelling east.

$d_1=2-v_1\cdot t$

$d_2=3-v_2\cdot t$

The distance between the two cars at any time ‘$t$’ can be written as

$d\left(t\right)=\sqrt{d_1^2+d_2^2}$

$d\left(t\right)=\sqrt{\left(4+v_1^2{t}^2-4v_{1}t\right)+\left(9+v_2^2{t}^2-6v_{2}t\right)}$

Substituting $v_1=40\text{ and}{v}_2=30$.

$=\sqrt{\left(4+900t^2-120t\right)+\left(9+1600t^2-240t\right)}$

$=\sqrt{\left(2500t^2-360\text{t}+13\right)}$

Therefore the distance between the cars at any time $t$ is $d\left(t\right)=\sqrt{\left(2500t^2-360\text{t}+13\right)}$.

To determine

To find:

b. Graph the function $d=d\left(t\right)$ and find the value of $t$ at which $d$ is smallest.

Answer to Problem 19AYU

Solution:

b. $t=0.07hours$

Explanation of Solution

Given:

Two cars are travelling perpendicular to each other with velocities 30 miles/hr and 40 miles/hr. One is 2 miles south of intersection and other is 3 miles east of intersection.

Calculation:

b. Graph the function $d=d\left(t\right)$ and find the value of $t$ at which $d$ is smallest.

From the graph it can be seen that when $t=0.7$, the distance between the cars ‘$d$’ attains its minimum value $0.22$. Therefore, $d$ is smallest when $t=0.7 \text{hours}$.